What is the extraneous solution to these equations? $\dfrac{x^2 + 50}{x - 5} = \dfrac{15x}{x - 5}$
Multiply both sides by $x - 5$ $ \dfrac{x^2 + 50}{x - 5} (x - 5) = \dfrac{15x}{x - 5} (x - 5)$ $ x^2 + 50 = 15x$ Subtract $15x$ from both sides: $ x^2 + 50 - (15x) = 15x - (15x)$ $ x^2 + 50 - 15x = 0$ Factor the expression: $ (x - 10)(x - 5) = 0$ Therefore $x = 10$ or $x = 5$ At $x = 5$ , the denominator of the original expression is 0. Since the expression is undefined at $x = 5$, it is an extraneous solution.